Answer:
49m/s
59.07 m/s
Step-by-step explanation:
Given that :
Distance (s) = 178 m
Acceleration due to gravity (a) = g(downward) = 9.8m/s²
Velocity (V) after 5 seconds ;
The initial velocity (u) = 0
Using the relation :
v = u + at
Where ; t = Time = 5 seconds ; a = 9.8m/s²
v = 0 + 9.8(5)
v = 0 + 49
V = 49 m/s
Hence, velocity after 5 seconds = 49m/s
b) How fast is the ball traveling when it hits the ground?
V² = u² + 2as
Where s = height = 178m
V² = 0 + 2(9.8)(178)
V² = 0 + 3488.8
V² = 3488.8
V = √3488.8
V = 59.07 m/s
Answer:
Numerator = 2(b^2+a^2) or equivalently 2b^2+2a^2
Denominator = (b+a)^2*(b-a), or equivalently b^3+ab^2-a^2b0-a^3
Step-by-step explanation:
Let
S = 2b/(b+a)^2 + 2a/(b^2-a^2) factor denominator
= 2b/(b+a)^2 + 2a/((b+a)(b-a)) factor denominators
= 1/(b+a) ( 2b/(b+a) + 2a/(b-a)) find common denominator
= 1/(b+a) ((2b*(b-a) + 2a*(b+a))/((b+a)(b-a)) expand
= 1/(b+a)(2b^2-2ab+2ab+2a^2)/((b+a)(b-a)) simplify & factor
= 2/(b+a)(b^2+a^2)/((b+a)(b-a)) simplify & rearrange
= 2(b^2+a^2)/((b+a)^2(b-a))
Numerator = 2(b^2+a^2) or equivalently 2b^2+2a^2
Denominator = (b+a)^2*(b-a), or equivalently b^3+ab^2-a^2b0-a^3
If this is a True or False question, that is true.
Answer:
= -21 - 15g
Step-by-step explanation:
Given that:
= -3(7+5g)
By simplifying
As "-" sign will invert the inner signs:
= -21 - 15g
I hope it will help you!!
